mauifan wrote:1) I believe that I am more concerned with output voltage right now. Since I don't believe that I need to drive anything with my circuit, I chose 1mA as the quiescent current.
Sounds reasonable. It's a good to get in the habit of adding a load though.. that's an important current path.
A large resistor is fine.. 100k to 1M.. but do throw it in. If you get used to thinking of the collector-emitter path in isolation, you pick up a set of expectations that get in the way once you actually try using transistor amplifiers.
mauifan wrote:2) Given that I am using Vcc=6V, I want the quiescent value of OUT- to be at 3V. Therefore I need a resistor Rc that will give me a 3V voltage drop at 1mA. Ohms Law: Rc = (Vcc - Vc) / Iq = (6V - 3V) / 1mA = 3k. As was the case in the op amp version of this thread, I used a potentiometer set at about 3k in my test circuit.
mauifan wrote:3) Although this step probably wasn't necessary given that Rc=3k, I chose Re=100 ohms.
It may not be necessary, but it's a really good idea.
It's almost impossible to design a bias network without an emitter resistor and get it to work. The tolerances of your components are larger than the acceptable error margin for Vbe. To keep Ic within 5% of its expected value, you have to be within 1.25mV of the correct Vbe. If you build your bias network from 5% resistors (expecting Vbe to be .6v), you'll actually get a voltage somewhere between .585v and .615v.. 24 times wider your tolerance for Vbe, and enough to change Ic by a factor of 3. 1% resistors give you a narrower window, but it's still about 8 times wider than you want, and large enough to move Ic 10% either side of the correct value.
The emitter resistor lets you ignore the details. You choose an approximate value for V.re and the transistor automatically floats to the Vbe necessary to send the appropriate current through. The exact value of V.re will probably be a few millivolts away from the predicted value, but the error will only be a small fraction of V.re.
mauifan wrote:The datasheet for the 9014 I am using indicates that its maximum collector current is 150mA. That means that it needs Rc to be a minimum of 40 ohms. The smallest resistor I have is 100 ohms. Given that I will likely vary the pot I used for Rc to test gain, I just figured that Re=100 ohms would help protect the transistor from myself. :D
That's good defensive design, but having an emitter resistor at all will prevent most problems. If you reduce R.c to zero, the circuit will end up being an emitter follower.. the voltage at the emitter will always stay about .6v below V.be.
mauifan wrote:4) Given that Re=100 is small compared to Rc=3k, I assumed that about 1mA would flow through Re in quiescent state. The voltage drop through Re is therefore Ve = Iq * Re = 1mA * 100 = 0.1V.
That's a good value.. small enough it won't consume a lot of your output range, but large enough for the 'V.be adjusting itself' errors to be a negligible fraction of V.re.
mauifan wrote:5) The datasheet doesn't mention it, but I assumed that the forward voltage drop between base and emitter is .6V. Adding in the drop across Re, the quiescent base voltage needs to be about Vb = 0.7V.
Yep. The actual equation for base voltage is:
I.b = I.s * e^(V.b/V.t)
where I.s is very small and e^(V.b/V.t) gets really big really fast.
I.s is called the 'intrinsic current' or 'reverse saturation current'. The details are complicated, but basically the thermal energy in a silicon crystal bumps some of the electrons up to energy levels where they can travel freely through the lattice. The boundaries between P and N doped semiconductors form regions that are almost perfect insulators, but the thermal effects allow a small amount of current to leak through. Diodes and BJTs work by manipulating that effect.
The actual leakage current (I.s) tends to be around 100 femtofarads (10^-13).
Bias voltage affects I.s exponentially, and when you get up around e^23 (10^10), the current approaches 1mV (10^-3).
.6v/.026v =~ 23.07, so that's a convenient estimate for V.b.
mauifan wrote:6) I now need to select a voltage divider R1 and R2 in a manner that results in output Vb = 0.7V. On top of that, R1 || R2 needs to be much less than the impedance looking into the base of Q1. If Q1 is fully on, the impedance of Rc || Re is very close to Re, so I assumed that the transistor has a gain of about 100 and obtained the base impedance at 100 * 100 = 10k.
You can ignore R.c entirely. The path between the collector and base is a diode, and the 'active region' where amplification happens is defined to be 'the range of voltages where the collector-base diode is reverse-biased'. If the collector-base diode gets forward biased (V.c is more than about .4v below V.b), it's called 'saturation'.
In other words, all the base should see is the path through the emitter. The equation for input resistance is:
beta * (R.e + 1/g.m)
where 1/g.m is about 26 ohms when I.c = 1mA. In this case, the total impedance should be about 12.6k. You were close enough for back-of-the-envelope work.
mauifan wrote:7) "Much Less" means that R1 || R2 needs to be about 1/10th of the base impedance, so I used a Thevenin equivalent power source of 0.7V with a 1K series resistor. My ElectroDroid calculator tells me that I need R1=4.7k, R2=1.6k, which I implemented in the form of a pot
Hmm.. I show that pair of resistors having a center voltage of about 1.5v when the supply is 6v. My number crunching says you want the upper resistor to be about 7-1/2 times as large as the lower one ((6-.7)/.7)
Figuring the parallel resistance for that ratio and scaling to 1k says you want 8.5k and 1.13k. Adjusting that for values in the E6 series gives:
- 6.8k + 680 over 1k : center voltage =~ .707v, parallel resistance =~ 880 ohms
- 10k + 1.5k over 1.5k : center voltage =~ .692v, parallel resistance =~ 1.3k
either of which should work.
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